31.(1+5^3+5^4+...+5^402) chia hết cho 31(dpcm)

gom: (1+5+5^2)+(5^3+5^4+5^5)+....(5^402+5^403+5^404)

=1(1+5+5^2)+5^3(1+5+5^2)+...+5^402(1+5+5^2)

=1.31+5^3.31+...+5^402.31

Vay 1+5+5^2+...+5^403+5^404chia het cho 31

B= 1+ 5+ 5^2+ 5^3+ ... + 5^96+ 5^97+ 5^98

=(1+5+5²)+(5³+5⁴+5⁵)+....+(5⁹⁶+5⁹⁷+5⁹⁸)

=31+(5³.1+5³.5+5³.5²)+....+(5⁹⁶.1+5⁹⁷.5+5⁹⁸.5²)

=31+5³.(1+5+5²)+....+5⁹⁶.(1+5+5²)

=31.1+5³.31+...+5⁹⁶.31

=31.(1+5³+...+5⁹⁶)

=> B chia hết cho 31

B = 1+5+5²+5³+....+5⁹⁸

B = (1+5+5²)+(5³+5⁴+5⁵)+....+(5⁹⁶+5⁹⁷+5⁹⁸)

B = 1(1+5+5²)+5³(1+5+5²)+....+5⁹⁶(1+5+5²)

B = 1.31 + 5³.31+.......+5⁹⁶.31

B = 31.(1+5³+.....+5⁹⁶) chia hết cho 31 (đpcm)

B = 1+5+52+53+....+598

B = (1+5+52)+(53+54+55)+....+(596+597+598)

B = 1(1+5+52)+53(1+5+52)+....+596(1+5+52)

B = 1.31 + 53.31+.......+596.31

B = 31.(1+53+.....+596) chia hết cho 31 (đpcm)

B = 5 + 5² + 5³ + ... + 5⁹⁰

= (5 + 5² + 5³) + (5⁴ + 5⁵ + 5⁶) + ... + (5⁸⁸ + 5⁸⁹ + 5⁹⁰)

= 5.(1 + 5 + 5²) + 5⁴.(1 + 5 + 5²) + ... + 5⁸⁸.(1 + 5 + 5²)

= 5.31 + 5⁴.31 + ... + 5⁸⁸.31

= 31.(5 + 5⁴ + ...+ 5⁸⁸) ⋮ 31

Vậy B ⋮ 31

\(B=5+5^2+5^3+...+5^{89}+5^{90}\)

Ta có: \(B=\left(5+5^2+5^3\right)+...+\left(5^{88}+5^{89}+5^{90}\right)\)

\(B=155+...+5^{87}.\left(5+5^2+5^3\right)\)

\(B=155+...+5^{87}.155\)

\(B=155.\left(1+...+5^{87}\right)\)

Vì \(155⋮31\) nên \(155.\left(1+...+5^{87}\right)⋮31\)

Vậy \(B⋮31\)

\(#WendyDang\)

Tớ nghĩ nên phải đổi số 5^4 thành 5^5

=(1+5+5^2)+...+5^402(1+5+5^2)

=31+...+5^402.31

=31(1+...+5^402) chia hết cho 31 

\(1+5+5^2+...+5^{404}=\left(1+5+5^2\right)+...+\left(5^{400}+5^{401}+5^{402}\right)=31+31.5^3+...+31.5^{400}\)

\(=31\left(1+5^3+5^6+...+5^{400}\right)\)chia hết cho 31